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adelina 88 [10]
2 years ago
8

3. Two people push on a car with a force of 5150 N, and the car moves a distance of 8.0m. What is the

Physics
2 answers:
MArishka [77]2 years ago
6 0
I think c not sure tho
tangare [24]2 years ago
3 0
The answer is D, work done is equal to force times it’s distance travelled, so 5150*8.0=41200J
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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At thi
tatiyna

Answer:

50 revolutions

Explanation:

Data provided:

case I: From rest to top spin

The initial angular speed of the washer, ωi = 0 rev /s

Final angular speed of the washer ωf = 5 rev /s

Time taken, t₁ = 8 s

now,  

The angular displacement or the number of revolutions taken (θ₁) is calculated as:

  θ₁ = ωi t₁ + (1/2)α₁t₁²

where,

α is the angular acceleration

The angular acceleration can be calculated as:

  ωf - ωi = α₁t₁

on substituting the values, we get

8α₁ = 5 - 0

or

α₁ = 0.625 rev/s²

substituting the values in the equation for the number of revolutions, we get

θ₁ = 0 + (1/2) (0.625)(8)²

or

θ₁ = 20 revolutions

also,  

For the case II: From top spin to rest

we have

The initial angular speed, ωi = 5 rev /s

and the final angular speed, ωf = 0 rev /s

Total time taken, t₂ = 12 s

Now, angular acceleration for this case

  ωf - ωi = α₂t₂

on substituting the values, we have

  12α₂ = 0 - 5

α₂ = - 0.4166 rev/s²

Therefore, the number of revolutions ( i.e angular displacement  )

θ₂ = ωit₂ + (1/2)α₂t₂²

on substituting the values, we have

θ₂ = 5 × 12 + (1/2)(-0.4166)(12)²

or

θ₂ = 30 rev

Hence,

the total number of revolutions made by the washer during the 20s is  

θ = θ₁ + θ₂

or

θ = 20 rev + 30 rev

or

θ = 50 revolutions

7 0
2 years ago
Your friend from France came to visit you when she was packing she went on weather.com and found that the average temperature in
Marina CMI [18]

Answer:

Because there is not as much cold as it was in France.

Explanation:

The average temperature in France during January ranges from 2.7° to 7.2° celsius which makes it the coldest month of the year. But since she comes to know that average temperature in Annville ranges 31° celsius which implies that the temperature is normal there and therefore, she packs sleeveless tops and shorts. Coats would not be required in a hot weather and hence, she does not pack it.

8 0
2 years ago
A(n) signal carries information by copying an original sound.
viva [34]
An analog signal carries information by copying an original sound
When we speak through the microphone, it turns our sound into some sort of electronic wave.
This electronic wave is caught by a recording device and later could be replicated into Mp3 file that we usually listen to
6 0
2 years ago
Read 2 more answers
A boxed 12.0 computer monitor is dragged by friction 7.50 up along the moving surface of a conveyor belt inclined at an angle of
sasho [114]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below is the solution:

W done by Normal = 0. (make the incline flat, Normal force goes directly up: no work done) 
<span>W done by gravity = w*displacement = (11kg*9.8) * 7.5sin(35) = -463J </span>
<span>W done by friction is the opposite of the work done by weight because the object is not moving. Therefore W done by friction = 463J</span>
8 0
3 years ago
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,
eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

x(t) = A cos (\omega t + \phi) (1)

where

A is the amplitude

\omega is the angular frequency

\phi is the phase

t is the time

The displacement of the piston in the problem is given by

x(t) = (5.00 cm) cos (5t+\frac{\pi}{5}) (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

v(t) = x'(t) = -\omega A sin (\omega t + \phi) (3)

Differentiating eq.(2), we find

v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})

And substituting t=0, we find the velocity at time t=0:

v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)

Differentiating eq.(3), we find

a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})

And substituting t=0, we find the acceleration at time t=0:

a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

\omega=\frac{2\pi}{T}

Using \omega = 5.0 rad/s and solving for T, we find

T=\frac{2\pi}{5 rad/s}=1.26 s

4 0
2 years ago
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